Problem: You know, it's hard to conduct a show with lots of participants and spectators at the same place nowadays. Still, you are not giving up on your dream to make a car crash showcase! You decided to replace the real cars with remote controlled ones, call the event "Remote Control Kaboom Show" and stream everything online.

For the preparation you arranged an arena — an infinite 2D-field. You also bought n remote controlled cars and set them up on the arena. Unfortunately, the cars you bought can only go forward without turning left, right or around. So you additionally put the cars in the direction you want them to go.

To be formal, for each car i (1 ≤ i ≤ n) you chose its initial position (x_i, y_i) and a direction vector (dx_i, dy_i). Moreover, each car has a constant speed s_i units per second. So after car i is launched, it stars moving from (x_i, y_i) in the direction (dx_i, dy_i) with constant speed s_i.

The goal of the show is to create a car collision as fast as possible! You noted that launching every car at the beginning of the show often fails to produce any collisions at all. Thus, you plan to launch the i-th car at some moment t_i. You haven't chosen t_i, that's yet to be decided. Note that it's not necessary for t_i to be integer and t_i is allowed to be equal to t_j for any i, j.

The show starts at time 0. The show ends when two cars i and j (i ≠ j) collide (i. e. come to the same coordinate at the same time). The duration of the show is the time between the start and the end.

What's the fastest crash you can arrange by choosing all t_i? If it's possible to arrange a crash then print the shortest possible duration of the show. Otherwise, report that it's impossible.

Input

The first line contains a single integer n (1 ≤ n ≤ 25000) — the number of cars.

Each of the next n lines contains five integers x_i, y_i, dx_i, dy_i, s_i (-10^3 ≤ x_i, y_i ≤ 10^3; 1 ≤ |dx_i| ≤ 10^3; 1 ≤ |dy_i| ≤ 10^3; 1 ≤ s_i ≤ 10^3) — the initial position of the i-th car, its direction vector and its speed, respectively.

It's guaranteed that all cars start at distinct positions (i. e. (x_i, y_i) ≠ (x_j, y_j) for i ≠ j).

Output

Print the shortest possible duration of the show if it's possible to arrange a crash by choosing all t_i. Otherwise, print "No show :(".

Your answer is considered correct if its absolute or relative error does not exceed 10^{-6}.

Formally, let your answer be a, and the jury's answer be b. Your answer is accepted if and only if \frac{|a - b|}{max{(1, |b|)}} ≤ 10^{-6}.

Examples

Input


4
3 -1 -1 1 2
2 3 -3 -2 10
-4 2 1 -2 1
-2 -2 -1 2 4


Output


0.585902082262898


Input


2
-1 1 -1 1 200
1 1 1 5 200


Output


No show :(

Note

Here is the picture for the first example: 

<image>

The fastest cars to crash are cars 2 and 4. Let's launch car 2 at 0, car 4 at about 0.096762 and cars 1 and 3 at arbitrary time. That way cars 2 and 4 will crash into each other at about 0.585902. So here's what it looks like at the moment of the collision:

<image>

Here's the picture for the second example:

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3fll;
const long double DINF = 1e18;
const long double pi = acos(-1.0);
const long double eps = 1e-9;
bool eq(long double a, long double b) { return abs(a - b) <= eps; }
struct pt {
  long double x, y;
  pt() {}
  pt(long double x_, long double y_) : x(x_), y(y_) {}
  bool operator<(const pt p) const {
    if (!eq(x, p.x)) return x < p.x;
    if (!eq(y, p.y)) return y < p.y;
    return 0;
  }
  bool operator==(const pt p) const { return eq(x, p.x) and eq(y, p.y); }
  pt operator+(const pt p) const { return pt(x + p.x, y + p.y); }
  pt operator-(const pt p) const { return pt(x - p.x, y - p.y); }
  pt operator*(const long double c) const { return pt(x * c, y * c); }
  pt operator/(const long double c) const { return pt(x / c, y / c); }
  long double operator*(const pt p) const { return x * p.x + y * p.y; }
  long double operator^(const pt p) const { return x * p.y - y * p.x; }
  friend istream& operator>>(istream& in, pt& p) { return in >> p.x >> p.y; }
};
struct line {
  pt p, q;
  line() {}
  line(pt p_, pt q_) : p(p_), q(q_) {}
  friend istream& operator>>(istream& in, line& r) { return in >> r.p >> r.q; }
};
long double dist(pt p, pt q) { return hypot(p.y - q.y, p.x - q.x); }
long double dist2(pt p, pt q) {
  return ((p.x - q.x) * (p.x - q.x)) + ((p.y - q.y) * (p.y - q.y));
}
long double norm(pt v) { return dist(pt(0, 0), v); }
long double angle(pt v) {
  long double ang = atan2(v.y, v.x);
  if (ang < 0) ang += 2 * pi;
  return ang;
}
long double sarea(pt p, pt q, pt r) { return ((q - p) ^ (r - q)) / 2; }
bool col(pt p, pt q, pt r) { return eq(sarea(p, q, r), 0); }
int paral(pt u, pt v) {
  if (!eq(u ^ v, 0)) return 0;
  if ((u.x > eps) == (v.x > eps) and (u.y > eps) == (v.y > eps)) return 1;
  return -1;
}
bool ccw(pt p, pt q, pt r) { return sarea(p, q, r) > eps; }
pt rotate(pt p, long double th) {
  return pt(p.x * cos(th) - p.y * sin(th), p.x * sin(th) + p.y * cos(th));
}
pt rotate90(pt p) { return pt(-p.y, p.x); }
bool isvert(line r) { return eq(r.p.x, r.q.x); }
long double getm(line r) {
  if (isvert(r)) return DINF;
  return (r.p.y - r.q.y) / (r.p.x - r.q.x);
}
long double getn(line r) {
  if (isvert(r)) return DINF;
  return r.p.y - getm(r) * r.p.x;
}
bool paraline(line r, line second) {
  return paral(r.p - r.q, second.p - second.q);
}
bool isinseg(pt p, line r) {
  if (p == r.p or p == r.q) return 1;
  return paral(p - r.p, p - r.q) == -1;
}
long double get_t(pt v, line r) { return (r.p ^ r.q) / ((r.p - r.q) ^ v); }
pt proj(pt p, line r) {
  if (r.p == r.q) return r.p;
  r.q = r.q - r.p;
  p = p - r.p;
  pt proj = r.q * ((p * r.q) / (r.q * r.q));
  return proj + r.p;
}
pt inter(line r, line second) {
  if (paraline(r, second)) return pt(DINF, DINF);
  if (isvert(r)) return pt(r.p.x, getm(second) * r.p.x + getn(second));
  if (isvert(second)) return pt(second.p.x, getm(r) * second.p.x + getn(r));
  long double x = (getn(second) - getn(r)) / (getm(r) - getm(second));
  return pt(x, getm(r) * x + getn(r));
}
bool interseg(line r, line second) {
  if (isinseg(r.p, second) or isinseg(r.q, second) or isinseg(second.p, r) or
      isinseg(second.q, r))
    return 1;
  return ccw(r.p, r.q, second.p) != ccw(r.p, r.q, second.q) and
         ccw(second.p, second.q, r.p) != ccw(second.p, second.q, r.q);
}
long double disttoline(pt p, line r) { return dist(p, proj(p, r)); }
long double disttoseg(pt p, line r) {
  if (isinseg(proj(p, r), r)) return disttoline(p, r);
  return min(dist(p, r.p), dist(p, r.q));
}
long double distseg(line a, line b) {
  if (interseg(a, b)) return 0;
  long double ret = DINF;
  ret = min(ret, disttoseg(a.p, b));
  ret = min(ret, disttoseg(a.q, b));
  ret = min(ret, disttoseg(b.p, a));
  ret = min(ret, disttoseg(b.q, a));
  return ret;
}
bool operator<(const line& a, const line& b) {
  if (a.p == b.p) return ccw(a.p, a.q, b.q);
  if (!eq(a.p.x, a.q.x) and (eq(b.p.x, b.q.x) or a.p.x + eps < b.p.x))
    return ccw(a.p, a.q, b.p);
  return ccw(a.p, b.q, b.p);
}
bool has_intersection(vector<line> v) {
  auto intersects = [&](pair<line, int> a, pair<line, int> b) {
    return interseg(a.first, b.first);
  };
  vector<pair<pt, pair<int, int>>> w;
  for (int i = 0; i < v.size(); i++) {
    if (v[i].q < v[i].p) swap(v[i].p, v[i].q);
    w.push_back({v[i].p, {0, i}});
    w.push_back({v[i].q, {1, i}});
  }
  sort(w.begin(), w.end());
  set<pair<line, int>> se;
  for (auto i : w) {
    line at = v[i.second.second];
    if (i.second.first == 0) {
      auto nxt = se.lower_bound({at, i.second.second});
      if (nxt != se.end() and intersects(*nxt, {at, i.second.second})) return 1;
      if (nxt != se.begin() and intersects(*(--nxt), {at, i.second.second}))
        return 1;
      se.insert({at, i.second.second});
    } else {
      auto nxt = se.upper_bound({at, i.second.second}), cur = nxt, prev = --cur;
      if (nxt != se.end() and prev != se.begin() and
          intersects(*nxt, *(--prev)))
        return 1;
      se.erase(cur);
    }
  }
  return 0;
}
int n;
vector<pair<pt, pt>> pts;
vector<int> S;
long double com = 0.218750000;
bool has = false;
bool show(long double m) {
  bool ret = false;
  vector<line> v;
  for (int i = 0; i < n; i++) {
    pt p = pts[i].first;
    pt q = pts[i].second;
    long double mul = m * S[i] / norm(q);
    q = q * mul;
    q = q + p;
    v.push_back(line(p, q));
  }
  ret = has_intersection(v);
  has |= ret;
  return ret;
}
long double bb(long double l, long double r) {
  if (eq(l, r)) return l;
  long double m = (l + r) / 2.0;
  if (show(m))
    r = m;
  else
    l = m;
  return bb(l, r);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    pt a, b;
    cin >> a >> b;
    int second;
    cin >> second;
    S.push_back(second);
    pts.push_back({a, b});
  }
  cout << setprecision(9) << fixed;
  long double ans = bb((long double)0, (long double)1e13);
  if (!has)
    cout << "No show :(" << '\n';
  else
    cout << ans << '\n';
  exit(0);
}